Trigonometric Identities and equations

AI Thread Summary
The discussion revolves around solving the trigonometric equation 5(sin x - cos x) = 4sin x - 3cos x within the domain 0° < x < 360°. Participants reference key trigonometric identities and attempt to simplify the equation, leading to sin x - 2 cos x = 0. Suggestions include rearranging the equation to isolate terms and dividing to facilitate solving. The conversation highlights the collaborative effort to find a solution, which was ultimately achieved. The focus remains on applying trigonometric identities to solve the equation effectively.
wei1006
Messages
6
Reaction score
0
1) Problem statement:
Solve the trigonometric equation for the domain is. 0°<x<360°
5(sinx - cosx) = 4sinx - 3cosx

2) Relevent equations:
secx=1/cosx
cosec x=1/sinx
cot x= 1/tanx
tan x=sinx/cosx
cot x=cosx/sinx
sin^2 x + cos^2 x=1
1 + tan^2 x= sec^2 x
1+ cot^2 x = cosec^2 x

Template of answer:
New domain:
Basic angle=
Quadrants=
Solve for x
3)Attempt the question
5(sin x - cosx)=4sinx - 3cosx
5sinx-4sinx-5cosx+3cosx=0
sinx - 2 cos x=0
 
Physics news on Phys.org
wei1006 said:
sinx - 2 cos x=0

Fine up to here, so what is the solution to this equation?
 
I am not sure how to continue...
 
How about trying to rewrite it by placing the 2 cos x on the other side and then dividing by something that makes it easier?
 
Thank you for your help, have solved it
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top